Circuits with Realistic Resistors
5.2 Terminal Voltages and Currents
Offset null 1
Inverting input
Noninverting input
NC
Output
Offset null 2
3
4
8
7
6
5
V
V
Figure 5.1 The eight-lead DIP package (top view).
1DIP is an abbreviation for dual in-line package. This means that the terminals on each side of the package are in line, and that the terminals on opposite sides of the package also line up.
2The common node is external to the op amp. It is the reference terminal of the circuit in which the op amp is embedded.
5.2 Terminal Voltages and Currents 147
Figure 5.5 shows the current variables with their reference directions.
Note that all the current reference directions are into the terminals of the operational amplifier: is the current into the inverting input terminal;
is the current into the noninverting input terminal; is the current into the output terminal; is the current into the positive power supply termi- nal; and is the current into the negative power supply terminal.
The terminal behavior of the op amp as a linear circuit element is characterized by constraints on the input voltages and the input currents.
The voltage constraint is derived from the voltage transfer characteristic of the op amp integrated circuit and is pictured in Fig. 5.6.
The voltage transfer characteristic describes how the output voltage varies as a function of the input voltages; that is, how voltage is transferred from the input to the output. Note that for the op amp, the output voltage is a function of the difference between the input voltages, The equation for the voltage transfer characteristic is
(5.1)
We see from Fig. 5.6 and Eq. 5.1 that the op amp has three distinct regions of operation. When the magnitude of the input voltage difference ( ) is small, the op amp behaves as a linear device, as the output voltage is a linear function of the input voltages. Outside this linear region, the output of the op amp saturates, and the op amp behaves as a nonlinear device, because the output voltage is no longer a linear function of the input voltages. When it is operating linearly, the op amp’s output voltage is equal to the difference in its input voltages times the multiplying constant, or gain
,
A.When we confine the op amp to its linear operating region, a con- straint is imposed on the input voltages, and The constraint is based on typical numerical values for and Ain Eq. 5.1. For most op amps, the recommended dc power supply voltages seldom exceed 20 V, and the gain,A, is rarely less than or We see from both Fig. 5.6 and Eq. 5.1 that in the linear region, the magnitude of the input voltage difference ( ) must be less than or 2 mV.
Typically, node voltages in the circuits we study are much larger than 2 mV, so a voltage difference of less than 2 mV means the two voltages are essentially equal. Thus, when an op amp is constrained to its linear operat- ing region and the node voltages are much larger than 2 mV, the constraint on the input voltages of the op amp is
(5.2)
Note that Eq. 5.2 characterizes the relationship between the input voltages for an ideal op amp; that is, an op amp whose value of Ais infinite.
The input voltage constraint in Eq. 5.2 is called the virtual short condition at the input of the op amp. It is natural to ask how the virtual short is maintained at the input of the op amp when the op amp is embedded in a circuit, thus ensuring linear operation. The answer is that a signal is fed back from the output terminal to the inverting input ter- minal. This configuration is known as negative feedback because the
vp = vn. 20>104,
|vp - vn|
104. 10,000,
VCC
vn. vp
|vp - vn| vo = c
- VCC A(vp - vn) 6 -VCC,
A(vp - vn) - VCC … A(vp - vn) … +VCC, + VCC A(vp - vn) 7 +VCC.
vp - vn. ic-
ic+
io
ip
in
VCC
V
V
VCC ip
io ic
in
ic
Figure 5.5 Terminal current variables.
Positive saturation
Negative saturation
(VCC /A) VCC
vo
(vp vn) Linear region
VCC
(VCC /A)
Figure 5.6 The voltage transfer characteristic of an op amp.
Input voltage constraint for ideal op amp
signal fed back from the output subtracts from the input signal. The negative feedback causes the input voltage difference to decrease.
Because the output voltage is proportional to the input voltage differ- ence, the output voltage is also decreased, and the op amp operates in its linear region.
If a circuit containing an op amp does not provide a negative feedback path from the op amp output to the inverting input, then the op amp will normally saturate. The difference in the input signals must be extremely small to prevent saturation with no negative feedback. But even if the cir- cuit provides a negative feedback path for the op amp, linear operation is not ensured. So how do we know whether the op amp is operating in its linear region?
The answer is, we don’t! We deal with this dilemma by assuming linear operation, performing the circuit analysis, and then checking our results for contradictions. For example, suppose we assume that an op amp in a circuit is operating in its linear region, and we compute the output voltage of the op amp to be 10 V. On examining the circuit, we discover that is 6 V, resulting in a contradiction, because the op amp’s output voltage can be no larger than Thus our assump- tion of linear operation was invalid, and the op amp output must be saturated at 6 V.
We have identified a constraint on the input voltages that is based on the voltage transfer characteristic of the op amp integrated circuit, the assumption that the op amp is restricted to its linear operating region and to typical values for and A. Equation 5.2 represents the voltage con- straint for an ideal op amp, that is, with a value of Athat is infinite.
We now turn our attention to the constraint on the input currents.
Analysis of the op amp integrated circuit reveals that the equivalent resist- ance seen by the input terminals of the op amp is very large, typically or more. Ideally, the equivalent input resistance is infinite, resulting in the current constraint
(5.3)
Note that the current constraint is not based on assuming the op amp is confined to its linear operating region as was the voltage constraint.
Together, Eqs. 5.2 and 5.3 form the constraints on terminal behavior that define our ideal op amp model.
From Kirchhoff’s current law we know that the sum of the currents entering the operational amplifier is zero, or
(5.4)
Substituting the constraint given by Eq. 5.3 into Eq. 5.4 gives
(5.5)
The significance of Eq. 5.5 is that, even though the current at the input terminals is negligible, there may still be appreciable current at the out- put terminal.
Before we start analyzing circuits containing op amps, let’s further sim- plify the circuit symbol. When we know that the amplifier is operating within its linear region, the dc voltages;VCCdo not enter into the circuit equations.
io = -(ic+ + ic-).
ip + in + io + ic+ + ic- = 0.
ip = in = 0.
1 MÆ VCC
VCC. VCC
Input current constraint for ideal op amp
5.2 Terminal Voltages and Currents 149
In this case, we can remove the power supply terminals from the symbol and the dc power supplies from the circuit, as shown in Fig. 5.7. A word of caution: Because the power supply terminals have been omitted, there is a danger of inferring from the symbol that We have already noted that such is not the case; that is, In other words, the ideal op amp model constraint that does not imply that
Note that the positive and negative power supply voltages do not have to be equal in magnitude. In the linear operating region, must lie between the two supply voltages. For example, if and , then . Be aware also that the value of A is not constant under all operating conditions. For now, however, we assume that it is. A discussion of how and why the value of Acan change must be delayed until after you have studied the electronic devices and components used to fabricate an amplifier.
Example 5.1 illustrates the judicious application of Eqs. 5.2 and 5.3.
When we use these equations to predict the behavior of a circuit contain- ing an op amp, in effect we are using an ideal model of the device.
-10 V … vo … 15 V V- = -10 V
V+ = 15 V vo
io = 0.
ip = in = 0 ip + in + io + ic+ + ic- = 0.
ip + in + io = 0.
vn
vp
vo
ip
io in
Figure 5.7 The op amp symbol with the power supply terminals removed.
Example 5.1 Analyzing an Op Amp Circuit
The op amp in the circuit shown in Fig. 5.8 is ideal.
a) Calculate if and
b) Repeat (a) for and
c) If specify the range of that avoids amplifier saturation.
Figure 5.8The circuit for Example 5.1.
Solution
a) Because a negative feedback path exists from the op amp’s output to its inverting input through the resistor, let’s assume the op amp is con- fined to its linear operating region. We can write a node-voltage equation at the inverting input terminal. The voltage at the inverting input termi- nal is 0, as from the connected volt- age source, and from the voltage constraint Eq. 5.2. The node-voltage equation at
is thus
i25 = i100 = in. vn
vn = vp vp = vb = 0 100 kÆ
25 k 10 V
10 V vo
va
vb
100 k
i25
i100
vb va = 1.5 V,
vb = 2 V.
va = 1 V
vb = 0 V.
va = 1 V vo
From Ohm’s law,
The current constraint requires Substituting the values for the three currents into the node-voltage equation, we obtain
Hence, is . Note that because lies between , the op amp is in its linear region of operation.
b) Using the same process as in (a), we get
Therefore, . Again, lies within .
c) As before, and Because
,
1.5 - vb
25 = -
vo - vb 100 . va = 1.5 V
i25 = -i100. vn = vp = vb,
;10 V vo
vo = 6 V i25 = -i100. i100 =
vo - vn
100 =
vo - 2 100 mA, i25 =
va - vn
25 =
1 - 2 25 = -
1 25 mA, vp = vb = vn = 2V,
; 10 V
vo
-4 V vo
1 25 +
vo
100 = 0.
in = 0.
i100 = (vo - vn)>100 = vo>100 mA.
i25 = (va - vn)>25 = 1 25 mA,
Objective 1—Use voltage and current constraints in an ideal op amp 5.1 Assume that the op amp in the circuit shown
is ideal.
a) Calculate for the following values of : 0.4, 2.0, 3.5, and
b) Specify the range of required to avoid amplifier saturation.
Answer: (a) 3, 8, and 10 V;
(b) -2 V … vs … 3 V.
-15, -10, -2,
vs
-2.4 V.
-1.6, -0.6,
vs
vo 16 k 10 V
15 V vo
vs
80 k
NOTE: Also try Chapter Problems 5.1, 5.4, and 5.5.
VCC
VCC
vo
is
in if
vp
vn
Rs
Rf
vs
Figure 5.9 An inverting-amplifier circuit.
Inverting-amplifier equation